Theory — Gibbs Energy Change \((\Delta G)\)
For processes that occur at constant temperature \(T\) and pressure \(P\) with only \(pV\) work,
spontaneity can be decided from the Gibbs energy change of the system:
\[
\begin{aligned}
\Delta G &= \Delta H - T\,\Delta S
\end{aligned}
\]
Here \(\Delta H\) is the enthalpy change (heat exchanged by the system at constant \(P\)),
and \(\Delta S\) is the entropy change of the system. The signs of \(\Delta G\) give the criterion:
- \(\Delta G < 0\): spontaneous
- \(\Delta G > 0\): nonspontaneous
- \(\Delta G = 0\): reversible / at equilibrium
Where does \(\Delta G\) come from?
The Second Law states that a process is spontaneous when the total entropy increases:
\[
\begin{aligned}
\Delta S_{\text{univ}}
&= \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \\
\Delta S_{\text{surr}} &= -\,\dfrac{\Delta H}{T} \\
T\,\Delta S_{\text{univ}}
&= T\,\Delta S_{\text{sys}} - \Delta H
\end{aligned}
\]
Define \(G = H - TS\). Then for changes of the system at constant \(T\):
\[
\begin{aligned}
\Delta G &= \Delta H - T\,\Delta S \\
&= -\,T\,\Delta S_{\text{univ}}
\end{aligned}
\]
Therefore, \(\Delta G < 0\) exactly corresponds to \(\Delta S_{\text{univ}} > 0\).
Temperature dependence and the threshold \(T^{*}\)
When both \(\Delta H\) and \(\Delta S\) are known, the temperature at which the balance flips
(\(\Delta G = 0\)) is
\[
\begin{aligned}
T^{*} &= \dfrac{\Delta H}{\Delta S}
\end{aligned}
\]
| Case | \(\Delta H\) | \(\Delta S\) | Result |
|---|
| 1 | \(<0\) | \(>0\) | Spontaneous at all \(T\) |
| 2 | \(>0\) | \(<0\) | Nonspontaneous at all \(T\) |
| 3 | \(>0\) | \(>0\) | Spontaneous at high \(T\) (\(T > T^{*}\)) |
| 4 | \(<0\) | \(<0\) | Spontaneous at low \(T\) (\(T < T^{*}\)) |
Units & conventions
- Use \(\mathrm{J\,mol^{-1}}\) (or \(\mathrm{kJ\,mol^{-1}}\)) for \(\Delta G\) and \(\Delta H\).
- Use \(\mathrm{J\,mol^{-1}\,K^{-1}}\) for \(\Delta S\).
- Temperatures must be in kelvin.
- Values are per mole of reaction as written (match the balanced equation).
Example: Vaporization of water at the normal boiling point
At \(T=373\ \mathrm{K}\): \(\Delta H \approx 40.7\ \mathrm{kJ\,mol^{-1}}\) and
\(\Delta S \approx 109\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
\[
\begin{aligned}
\Delta G &= \Delta H - T\,\Delta S \\
&= 40.7\times 10^3 - 373\times 109 \\
&\approx 0\ \mathrm{J\,mol^{-1}}
\end{aligned}
\]
So boiling at the normal boiling point is reversible/at equilibrium (\(\Delta G=0\)).
Using this calculator
- Choose what to solve for: \(\Delta G\), \(T\), or \(\Delta S\).
- Enter the remaining known quantities (and make sure units are consistent).
- Read the result and the spontaneity verdict; if desired, inspect the computed
threshold \(T^{*}=\Delta H/\Delta S\).
Beyond scope: standard-state thermodynamics links \(\Delta G^\circ\) with the equilibrium constant via
\(\Delta G^\circ = -RT\ln K\).
